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Questions
Heat energy is supplied at a constant rate to 100g of ice at 0 °C. The ice is converted into water at 0° C in 2 minutes. How much time will be required to raise the temperature of water from 0 °C to 20 °C? [Given: sp. heat capacity of water = 4.2 J g-1 °C-1, sp. latent heat of ice = 336 J g-1].
100 g of ice at 0°C receives a steady flow of heat energy. At 0°C, the ice melts into the water in 2 minutes. How long will it take to bring the temperature of the water up to 20 degrees? (Specific latent heat of ice = 336 J/g, Specific heat capacity of water = 4.2 J/g °C).
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Solution
Heat energy required to melt 100 g of ice at 0 °C is
Q = mL = 100 × 336=33600 J
Therefore, heat energy supplied per minute is
`33600/2=16800 "J min"^-1`
Now, heat energy required to raise the temperature of water from 0°C to 20°C is
Q'=mc × rise in temperature =100 × 4.2 × 20 = 8400 J
If time required for this heat gain is t minutes, then Heat energy supplied in t minutes is
`16800 xx "t"=8400`
`therefore "t" =8400/16800`
`"t" = 0.5 "min"`
`"t" = 30 "s"`
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